Statistical Manifold

Statistical Manifold

Amari-Chentsov Tensor

Given a pair of conjugate connections $\nabla$ and $\tilde{\nabla }$ on $(M,g)$, the Amari-Chentsov tensor is a $(0,3)$-tensor defined by$C_{ijk}:={\Gamma }_{ijk}-{\tilde{\Gamma }}_{ijk}$or in coordinate-free form, $C(X,Y,Z):=⟨{\nabla }_{X}Y-{\tilde{\nabla }}_{X}Y,Z⟩$

Proposition

The Amari-Chentsov tensor is totally symmetric if the conjugate connections are torsion-free.

Proof Since both connections are torsion-free, it is easy to show that $C(X,Y,Z)=C(Y,X,Z)$. By conjugation, we have $Xg(Y,Z)=g({\nabla }_{X}Y,Z)+g(Y,{\bar{\nabla }}_{X}Z),\text{ for all }X,Y,Z\in \mathcal{X}(M)$ Now consider $\begin{array}{rl}C(X,Z,Y)& =⟨{\nabla }_{X}Z-{\bar{\nabla }}_{X}Z,Y⟩\\ & =⟨{\nabla }_{X}Z,Y⟩-⟨{\bar{\nabla }}_{X}Z,Y⟩\\ & =X⟨Y,Z⟩-⟨Z,{\bar{\nabla }}_{X}Y⟩-X⟨Z,Y⟩+⟨{\nabla }_{X}Y,Z⟩\\ & =⟨{\nabla }_{X}Y-{\bar{\nabla }}_{X}Y,Z⟩\\ & =C(X,Y,Z)\end{array}$Therefore since any symmetric group is generated by adjacent transpositions, the Amari-Chentsov tensor is totally symmetric. $\square$

Statistical Manifold

A statistical manifold $(M,g,\nabla ,\tilde{\nabla },C)$ is a manifold $M$ equipped with a metric tensor $g$, a pair of torsion-free conjugate connections $(\nabla ,\tilde{\nabla })$ and corresponding totally symmetric Amari-Chentsov $C$.

$\alpha$-Connections

$\alpha$-Connection

For any statistical manifold $(M,g,\nabla ,\tilde{\nabla },C)$, we can define a family of connections $\{{\nabla }^{\alpha }{\}}_{\alpha \in \mathbb{R}}$ called $\alpha$-connections, such that $g({\nabla }_X^{\alpha }Y,Z)=g({\nabla }_X^{\mathrm{LC}}Y,Z)+\frac{\alpha }{2}C(X,Y,Z)\text{ for all }X,Y,Z\in \mathfrak{X}(M)$

Proposition

${\nabla }^{\alpha }$ is torsion free for any $\alpha \in \mathbb{R}$.

Proof Choose some coordinate chart, then we can derive the coordinate form: $\begin{array}{r}{\Gamma }_{ij,k}^{\alpha }={\Gamma }_{ij,k}^{\mathrm{LC}}-\frac{\alpha }{2}{C}_{ij,k}\\ {\Gamma }_{ij,k}^{-\alpha }={\Gamma }_{ij,k}^{\mathrm{LC}}+\frac{\alpha }{2}{C}_{ij,k}\end{array}$where ${\Gamma }_{ki,j}={\sum }_l{\Gamma }_{ij}^l{g}_{lk}$. Observe that they don’t change if swap indices $i$ and $j$, thus they are torsion free by theorem. $\square$

Proposition

For any fixed $\alpha \in \mathbb{R}$, $({\nabla }^{\alpha },{\nabla }^{-\alpha })$ forms a pair of conjugate connection.

Proof $\begin{array}{rl}& ⟨{\nabla }_X^{\alpha }Y,Z⟩+⟨y,{\nabla }_X^{-\alpha }Z⟩\\ =& ⟨{\nabla }_X^{\mathrm{LC}}Y,Z⟩+\frac{\alpha }{2}C(X,Y,Z)+⟨{\nabla }_X^{\mathrm{LC}}Z,Y⟩-\frac{\alpha }{2}C(Y,X,Z)\\ =& ⟨{\nabla }_X^{\mathrm{LC}}Y,Z⟩+⟨Y,{\nabla }_X^{\mathrm{LC}}Z⟩\\ =& X⟨Y,Z⟩\end{array}$The last equation holds since the Levi-Civita connection is self-conjugate. $\square$

Therefore for any $\alpha \in \mathbb{R}$, $(M,g,{\nabla }^{\alpha },{\nabla }^{-\alpha })$ forms a statistical manifold with totally symmetric cubic tensor $C$ being defined correspondingly.

Proposition

The $\alpha$-connections ${\nabla }^{\alpha }$ can also be constructed directly from a pair $(\nabla ,\tilde{\nabla })$ of conjugate connections by taking the following weighted combination: ${\nabla }^{\alpha }=\frac{1+\alpha }{2}\nabla +\frac{1-\alpha }{2}\tilde{\nabla }$